1. Field of the Invention
The present invention relates to a technique for high-accuracy driving of a swing oscillation element in an optical scanning apparatus and an image forming apparatus.
2. Description of the Related Art
Conventionally, various optical deflection apparatuses have been proposed where a mirror is resonance-driven. For example, Japanese Patent Laid-Open No. 2005-292627 discloses a resonant type optical deflection apparatus based on a single resonance frequency. The resonant type optical deflection apparatus based on a single resonance frequency has a problem in that the scanning angle of the mirror varies in a sine wave form in principle, resulting in a problem in that the angular velocity is not constant. In order to solve this, Japanese Patent Laid-Open No. 2005-326462 proposes a resonant type optical deflection apparatus based on an approximately-constant angular velocity. With the resonant type optical deflection apparatus disclosed in Japanese Patent Laid-Open No. 2005-326462, it is possible to generate a swing wave having a saw-tooth waveform.
FIG. 10 is an example block diagram illustrating the configuration of a control system for performing track control of an approximately-constant velocity resonant type optical deflection apparatus that has, as its natural frequencies, a fundamental frequency and a second harmonic thereof. A scanner block 101 of the approximately-constant velocity resonant type optical deflection apparatus is driven as described below. First, in accordance with a sequence control by a CPU 102, an NCO (numerically controlled oscillator) 201 generates a driving wave of the resonant type optical deflection apparatus. The driving wave output from the NCO 201 is subjected to PWM with a PWM modulator 104, and an H-bridge 105 is switched by the output from the PWM modulator 104, thereby performing linear modulation. The output from the H-bridge 105 drives the scanner block 101 of the resonant type optical deflection apparatus. The scanner block 101 has two oscillation modes, namely, a fundamental frequency and a second harmonic thereof. Therefore, the track with an approximately-constant angular velocity is defined by establishing the track equation stated below, the equation having four independent variables as coefficients, which are a combination of the vector of the fundamental frequency (amplitude A1 and phase Φ1) and the vector of the second harmonic (amplitude A2 and phase Φ2).θ=A1 sin(ωt+Φ1)+A2 sin(2ωt+Φ2)  Equation (1)
Four control loops composed of four-dimensional orthogonal component data are converged independent of each other in a linear control system that achieves the track defined by the above track equation, since the linear system is required to be mathematically holomorphic. The feedback loop shown in FIG. 10 starts from optical beam detection sensors 405 and 406, passes detected beam phases 1106/target beam phases 1107, a comparator 110, a conversion matrix 111, a PID system 114, and a feedback gain 115 in this order, and reaches the NCO 201. Here, the detected beam phases 1106 are indicated by counters P1, P2, P3, P4, and the target beam phases 1107 are indicated by P10; P20; P30; P40. This feedback loop is composed of a vector that includes four independent variables as its components. The optical beam detection sensors 405 and 406 are disposed in positions in a non-image region, the positions being substantially symmetrically with respect to, and located on the left and right sides of a center of swing. The optical beam detection sensors 405 and 406 each detect scanning track information that is component-displayed by four detected beam phases, P1, P2, P3, P4, of an optical beam that is deflected by a swing mirror 301 and travels back and forth.
The comparator 110 subtracts, from the detected beam phases 1106, the target beam phases 1107 of four points at the optical beam detection positions, the phases having been obtained in advance with the track equation, and outputs a beam phase difference vector. The beam phase difference vector is subjected to coordinate conversion by the conversion matrix 111 into difference control parameters composed of an amplitude component and a phase component, and the difference control parameters are input to a classical control theory PI control unit (PID system 114). Then, feedback amounts, ΔA1; ΔA2; Δφ1; Δφ2, obtained by multiplying the input difference control parameters by a feedback gain 115 are reflected in the control parameters of the track equation, thereby forming a feedback control unit. Here, a conversion matrix M−1 is obtained by performing partial differentiation on a target track equation with respect to each of eight components, namely, four orthogonal vector components of an input vector X and those of an output vector Y, solving a differential coefficient ratio of each output component to the corresponding input component, and generating a matrix. The matrix is a linear conversion matrix, and also a coefficient matrix for performing coordinate conversion from a space in the vector X into a space in the vector Y by a linear conversion operation, and thus the relevant equation needs to be a linear function in order to obtain an exact solution. However, the sine function in the track equation is a non-linear function involving infinite orders and is a composite function thereof, and thus it is impossible to directly solve an inverse function. Therefore, in order to calculate the conversion matrix M−1, a target track equation is subjected to Taylor expansion with respect to an optical beam detection point, so as to convert the equation into a linear approximate equation. At the same time, by taking the input vector X as a control parameter and the output vector Y as a beam phase, a conversion coefficient ratio matrix from the vector X to the vector Y is calculated once. Then, an inverse matrix thereof is obtained. In this manner, the conversion matrix M−1 is calculated in two steps. The concept of these two steps is described with the equations stated below.
                                                                        Δ                ⁢                                                                  ⁢                Y                            =                            ⁢                                                                    [                    M                    ]                                    ⁢                  Δ                  ⁢                                                                          ⁢                  X                                =                                                                                                      ⁢                                                                    [                                                                                            ∂                          Θ                                                                          ∂                          X                                                                    ⁢                                              |                                                  ϕ                          =                                                      Pn                            ⁢                                                                                                                  ⁢                            0                                                                                                                ]                                    ⁢                  Δ                  ⁢                                                                          ⁢                  X                                ⁢                                  →                                      INVERSE                    ⁢                                                                                  ⁢                    FUNCTION                                                  ⁢                                                                            [                      M                      ]                                                              -                      1                                                        ⁢                  Δ                  ⁢                                                                          ⁢                  Y                                                                                        Equation        ⁢                                  ⁢                  (          2          )                                                              Δ            ⁢                                                  ⁢            Y                    =                                    (                                                                                          Δ                      ⁢                                                                                          ⁢                      P                      ⁢                                                                                          ⁢                      1                                                                                                                                  Δ                      ⁢                                                                                          ⁢                      P                      ⁢                                                                                          ⁢                      2                                                                                                                                  Δ                      ⁢                                                                                          ⁢                      P                      ⁢                                                                                          ⁢                      3                                                                                                                                  Δ                      ⁢                                                                                          ⁢                      P                      ⁢                                                                                          ⁢                      4                                                                                  )                        =                          (                                                                                                                  P                        ⁢                                                                                                  ⁢                        1                                            -                                              P                        ⁢                                                                                                  ⁢                        10                                                                                                                                                                                                          P                          ⁢                                                                                                          ⁢                          2                                                -                                                  P                          ⁢                                                                                                          ⁢                          20                                                                    ⁢                                                                                                                                                                                                                              P                        ⁢                                                                                                  ⁢                        3                                            -                                              P                        ⁢                                                                                                  ⁢                        30                                                                                                                                                                                P                        ⁢                                                                                                  ⁢                        4                                            -                                              P                        ⁢                                                                                                  ⁢                        40                                                                                                        )                                      ⁢                                  ⁢                              Δ            ⁢                                                  ⁢            X                    =                                    (                                                                                          Δ                      ⁢                                                                                          ⁢                      A                      ⁢                                                                                          ⁢                      1                                                                                                                                  Δ                      ⁢                                                                                          ⁢                      A                      ⁢                                                                                          ⁢                      2                                                                                                                                  Δ                      ⁢                                                                                          ⁢                      ϕ                      ⁢                                                                                          ⁢                      1                                                                                                                                  Δ                      ⁢                                                                                          ⁢                      ϕ                      ⁢                                                                                          ⁢                      2                                                                                  )                        =                          (                                                                                                                  A                        ⁢                                                                                                  ⁢                        1                                            -                                              A                        ⁢                                                                                                  ⁢                        10                                                                                                                                                                                A                        ⁢                                                                                                  ⁢                        2                                            -                                              A                        ⁢                                                                                                  ⁢                        20                                                                                                                                                                                ϕ                        ⁢                                                                                                  ⁢                        1                                            -                                              ϕ                        ⁢                                                                                                  ⁢                        10                                                                                                                                                                                ϕ                        ⁢                                                                                                  ⁢                        2                                            -                                              ϕ                        ⁢                                                                                                  ⁢                        20                                                                                                        )                                                          Equation        ⁢                                  ⁢                  (          3          )                    
Here, Pn0 indicates a design target phase at a BD detection position.
Since the conversion matrix M−1 obtained by the above operation is a linear approximation, occurrence of an error is more likely as the distance from the expansion point increases. However, if a target beam phase that is determined depending on a design value of the optical beam detection position is used as the convergence point, in the vicinity thereof, the conversion matrix M−1 shows sufficient accuracy for practical use and thus can be used. With the configuration of the control system described above, the scanning track information composed of the four phase components P1; P2; P3; P4 detected by the optical beam detection sensors 405 and 406 are compared with a target track represented by the phase components P10; P20; P30; P40. Error information is converted with the conversion matrix M−1 into four control parameters composed of amplitude components and phase components, namely, ΔA1; ΔA2; Δφ1; Δφ2, which can be input to the feedback control unit, thereby realizing feedback control in a vector quantity. In this manner, a track control method for an approximately constant velocity resonant type optical deflection unit having a second harmonic as its natural frequency is proposed, in which the track is converged into a target track having four parameters.
However, in the track control of a resonant type optical deflection apparatus that has a second harmonic as its natural frequency, an offset may occur, the offset being a deviation amount in the scanning angle between a reference detection position of the optical beam detection sensor and a center of swing of the optical deflection apparatus. That is, there is a problem that if track control is performed assuming that the offset amount is zero, an offset appears as a phase control error of the second harmonic. The track equation including an offset amount is as described below. Note that the offset amount will be described in an embodiment described later with reference to FIG. 4.θ=A1 sin(ωt+Φ1)+A2 sin(2ωt+Φ2)+Offset  Equation (4)
That is, since the track equation is based on a sine function, in principle, the gain of the variation of the detection timing in the phase direction with respect to the variation in the amplitude direction, that is, the detection position angle, is at least one, and the gain increases as the distance to the peak decreases. Since the detection position of the optical beam detection sensor is set avoiding an image rendering region, the detection position is set on the side close to the peak of the scanning track, and thus the gain increases. Then, with respect to the second harmonic component, the detection phase is varied in the same direction as the variation of the amplitude, due to its phase relation with the fundamental frequency, and the amplitude is smaller than that of the fundamental frequency. Thus, the gain with respect to the phase deviation of the second harmonic component becomes higher.
FIG. 11 shows track velocity error due to an offset, and shows track error due to a phase deviation of a second harmonic that has occurred as a result of performing control with an offset of 1/(60)° as a graph. In the graph, the horizontal axis indicates the scanning angle and the vertical axis indicates the scanning speed. A waveform 1301 is a scanning speed waveform of the track affected by a phase deviation of the second harmonic, and the range between waveforms 1302 and 1303 is the acceptable range of a normal scanning speed waveform. In this manner, control error due to an offset error is conspicuously reflected, and a phase deviation of the second harmonic distorts the approximately-constant velocity track, and therefore, a scanning error due to the distortion occurs. Because of the reason described above, there is a problem that even a small offset amount causes a phase deviation of the second harmonic, and results in a very conspicuous scanning error as shown in FIG. 11, such that the image is distorted. Also, since the track deviates from the convergence point for the target track, a sufficient feedback gain cannot be applied, and jitter increases. However, in the case where a linear control system is designed in which an offset item is added to the target track equation and offset compensation is performed based on five-variance control, it is necessary to increase the number of phase components detected by the optical beam detection sensor from four to five. Therefore, it is necessary to add one more detection position, which is not preferable in terms of cost.